which a `U^(238)` nucleus original at rest , decay by emitting an alpha particle having a speed `u` , the recoil speed of the residual nucleus is

Explanation

To solve the problem of finding the recoil speed of the residual nucleus after a uranium-238 nucleus emits an alpha particle, we can follow these steps:

### Step 1: Understand the decay process
The uranium-238 nucleus (U-238) is initially at rest. When it decays, it emits an alpha particle (which is a helium nucleus with an atomic mass of 4 and atomic number of 2) and leaves behind a residual nucleus (thorium-234, with an atomic mass of 234 and atomic number of 90).

### Step 2: Apply conservation of momentum
Since the uranium nucleus is initially at rest, the total momentum before the decay is zero. According to the law of conservation of momentum, the total momentum after the decay must also equal zero.

Let:
- \( m_1 = 4 \) (mass of the alpha particle)
- \( u \) = speed of the alpha particle
- \( m_2 = 234 \) (mass of the residual nucleus, thorium-234)
- \( v \) = recoil speed of the residual nucleus

The momentum before decay is:
\[ 0 \]

The momentum after decay is:
\[ m_1 \cdot u + m_2 \cdot (-v) = 0 \]

### Step 3: Set up the equation
From the conservation of momentum, we can write:
\[ 4u - 234v = 0 \]

### Step 4: Solve for the recoil speed \( v \)
Rearranging the equation gives:
\[ 4u = 234v \]
\[ v = \frac{4u}{234} \]

### Step 5: Simplify the expression
We can simplify this further:
\[ v = \frac{2u}{117} \]

### Final Answer
The recoil speed of the residual nucleus (thorium-234) is:
\[ v = \frac{2u}{117} \]

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